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Inessa05 [86]
3 years ago
12

12. A runner at the start of a race generates 250 W of power as he accelerates to 5 m/s. If the runner has a mass of

Physics
1 answer:
Margaret [11]3 years ago
4 0
1/2 m/s2
30 Na




I hope this helps!
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What is true about valence electrons?
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The valence electrons are the one furthest from the nucleus
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3 years ago
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The figure shows a 1 cm cube that has a mass of 1 gram. Calculate the density of the cube in kg/m³.
dolphi86 [110]
Density = 1 g/cm^3 = 1000 Kg/m^3
3 0
3 years ago
For a photoelectric tube, calculate the voltage which will be just sufficient to stop electrons emitted by the sodium photo-plat
Talja [164]

Answer:

1.11 V

Explanation:

Given that the Einstein photoelectric equation states that;

KE = E - Wo

E = energy of incident photon

Wo= work function of the metal

E = hf = 6.64 x 10-34 * 6 x 1014

E = 39.84 * 10^-20 J or 3.98  * 10^-19 J

KE = 3.98  * 10^-19 J - 2.2 x 10-19J

KE = 1.78 *  10^-19J

We convert this value of KE to electron volts

KE = 1.78 *  10^-19J/1.6 x 10-19C

KE = 1.11 eV

Hence; 1.11 V will be just sufficient to stop electrons emitted by the sodium photo-plate reaching the collector plate.

8 0
3 years ago
How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five
Keith_Richards [23]

Answer:

<em>The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

<em></em>

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = I

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = 5I  (five times larger)

angular speed = ω/5  (five times smaller)

Recall that the kinetic energy of a spinning body is given as

KE = \frac{1}{2}Iw^{2}

therefore,

for the first case, the K.E. is given as

KE = \frac{1}{2}Iw^{2}

and for the second case, the K.E. is given as

KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2}   = \frac{5}{50}Iw^{2}

KE = \frac{1}{10}Iw^{2}

<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>

<em>This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

6 0
3 years ago
The plain flight covers 400 miles in the first hour. As you approach the landing site, it takes you an hour to cover the last 20
tatyana61 [14]

that is an example of negative acceleration because it is slowing down

3 0
3 years ago
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